JEE MAIN - Physics (2018 (Offline) - No. 10)
A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material
of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be :
0.9 n C
1.2 n C
0.3 n C
2.4 n C
Explanation
Charge on Capacitor initially,
Qi = CV
After inserting dielectric of dielectric constant = K,
new capacitance, Qf = (KC) $$ \vee $$
$$\therefore\,\,\,$$ Induced charges on dielectric
= Qf $$-$$ Qi
= KCV $$-$$ CV
= (K $$-$$ ) CV
= $$\left( {{5 \over 3} - 1} \right)$$ $$ \times $$ 90 $$ \times $$ 10$$-$$12 $$ \times $$ 20
= 1.2 $$ \times $$ 10$$-$$9 C
= 1.2 nC
Qi = CV
After inserting dielectric of dielectric constant = K,
new capacitance, Qf = (KC) $$ \vee $$
$$\therefore\,\,\,$$ Induced charges on dielectric
= Qf $$-$$ Qi
= KCV $$-$$ CV
= (K $$-$$ ) CV
= $$\left( {{5 \over 3} - 1} \right)$$ $$ \times $$ 90 $$ \times $$ 10$$-$$12 $$ \times $$ 20
= 1.2 $$ \times $$ 10$$-$$9 C
= 1.2 nC
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